I have a smooth plane curve $\gamma$ enclosed in a circle of radius $R$. See Figure 1. I'm interested in measuring the deviation of this curve from the $x$ axis, denoted $\Delta(t)$ in the figure, as a function of the curvature. In particular I would like to bound $\max_{t} {\Delta(t)}$. I know the curvature of $\gamma$ is bounded by the curvature of the circle. Thus $\kappa(t) < \frac{1}{R}$. I originally tried to solve this by summing up the curvature along $\gamma$ but that gave $\int_{0}^{\Delta t} \kappa(t) dt < \int_{0}^{\Delta t} \frac{1}{R} dt = \frac{\Delta t}{R}$, which doesn't seem right.
EDIT: I'm trying to bound $\max_{t}\Delta(t)$ as a function of the curvature $\kappa(t)$. Let $\tilde{\kappa} = \max_{t}{\kappa(t)}$. Then $\max_{t}{\Delta(t)}(\tilde{\kappa}) = 0$ when $\tilde{\kappa} = 0$, and $\gamma$ is a straight line (assume in this case the $\dot{\gamma} = \hat{i}$), and $\max_{t}{\Delta(t)}(\tilde{\kappa}) = R$ when $\tilde{\kappa} = \frac{1}{R}$, and $\gamma$ lies along the circle. What is the relationship for $\max_{t}{\Delta(t)}(\tilde{\kappa})$ for all other values of $\tilde{\kappa}$?
It is key that $\gamma$ is forced to pass through the black points on the sides of the circle shown in Figure 1.
